Experiment No.:
To Determine the Velocity Constant for the Hydrolysis of Methylacetate Using an
Acid as Catalyst at Laboratory Temperature.
Requirement: Methylacetate, N/2 HCl solution, N/15 NaOH solution, Phenolphthalein
(indicator), Ice cold water, 2ml pipette, Conical flask, Conical flask with stopper
(Erlenmeyer flask), 50ml measuring cylinder, 150ml & 250ml beaker.
Theory: Methylacetate hydrolyses in presence of an acid (acts as catalyst) and produces
acetic acid and methyl alcohol.
CH3COOCH3 +
H2O
H+
CH3COOH + CH3OH
Velocity of the reaction, v  dx  k ' CH 3 COOCH 3 H 2 O
dt
 k CH 3COOCH3 
where, k  k ' H 2 O
(Note: water is present in large excess, water concentration practically remain constant
throughout the reaction.)
In the above equation rate of reaction is determined by the first power of the concentration
(of ester). Hence, the reaction is called first order reaction. It is also known as pseudounimolecular reaction.
Let, a = initial concentration of ester (at t = 0) of the above reaction.
(a-x) = Concentration of ester remaining at time t.
The rate is represented by
dx
v
 k a  x 
dt
2.303
a
On integration, k 
log
t
( a  x)
This reaction was followed by quenching an aliquot of the reaction mixture at different
time and titrating the total acid (both acetic acid formed and the N/2 HCl acid taken
originally) by a standard solution of N/15 NaOH.
If V0, Vt and V are the volumes of N/15 NaOH consumed to neutralize the acids of a
given volume of the reaction mixture at the beginning (t = 0), at time t, and at the end
( t = ) of the reaction respectively, the rate constant of the reaction can be expressed as
k
V  V0
2.303
log 
t
V  Vt
---(1)
where V0= titration reading at time t=0, at time t, and at t = .
Prepared by Dr. S. Saha, Chemistry, BHU
1
because V  V0  is proportional to the concentration of the total acetic acid formed or
the conc. of ester in the beginning of the reaction, a and V  Vt  is proportional to the
concentration of the unreacted ester (a-x) at time t. The equation 1 can be rearranged as
follows
log
V  V0
k

t
V  Vt 2.303
[Note the similarity with y = mx + c (=0)]
The velocity constant was determined from the slope (m =
k
) of the graph of
2.303
V  V0
vs t as well as directly using the equation 1 at each time.
V  Vt
[An example of the graph was given at the end of this note.]
log
Procedure: N/15 NaOH solution was taken in the burette while 50ml N/2 HCl solution
was taken in a stoppered conical flask using a measuring cylinder. 2ml methylacetate was
pipetted out and poured into the conical flask and was shaked well. As soon as the ester
was poured, the time count starts. 2ml of the reaction mixture was pipetted out
immediately and was added to 25 ml of ice cold water in a conical flask (i.e., quenching
of the reaction). 2-3 drops of phenolphthalein indicator was added in that quenched
solution and was titrated against N/15 NaOH solution. Titration was completed as soon as
possible. The burette reading, so obtained, was considered as V0 (since t = 0 min,
assumed). When the stopwatch was seen to be at 5 min, again 2 ml of the reaction
mixture was quenched and the titration was done following the same procedure. The
burette reading so obtained was considered as V5 (since t = 5 min). The same procedure
was repeated at t = 10, 20, 30, and 40 min to get the corresponding burette readings of
V10, V20, V30, and V40 respectively [Note: Dany other convenient time gap may be used].
Finally, the reaction mixture was heated at 60-80o C for 20min (caution: Do not boil the
solution. Do not stoppered the conical tightly, use a paper strip to avoid stuck). It was
cooled down to room temperature. 2ml of this mixture was taken out in the conical and
was titrated against the N/15 NaOH as described earlier. The reading was noted as V.
Prepared by Dr. S. Saha, Chemistry, BHU
2
Observation and Calculation:
Lab. Temperature:
V0 =
ml
V =
ml
ml
V  V0  =
o
C
Table. Burette reading at different times and corresponding calculated velocity constants.
Sl.
no.
Time/
min
1.
2.
3.
4.
5.
5
10
20
30
40
Burette
reading
V  Vt 
V  V0
V  Vt
log
V  V0
V  Vt
k
V  V0
2.303
min-1
log 
t
V  Vt
From the Table : k avg =
From the Graph :
k Graph = slope x 2.303 min-1
Result: Velocity constant for the hydrolysis of the methylacetate at laboratory
temperature ( oC) was found to be:
i) From table, k avg =
min-1
min-1
ii) From Graph, k Graph =
Experiment No.:
To determine the velocity constant and half-life for the hydrolysis of methylacetate
acid as catalyst at laboratory temperature.
At t= t/2 (i.e., half life period: half of the reactant converted to the product)
x= a/2
k
2.303
a
2.303
log

log
t
( a  x)
t1
2
t 1  0.693
2
a
a
a
2

2.303
log 2
t1
2
0.693
min 1
k
Prepared by Dr. S. Saha, Chemistry, BHU
3
Example Graph 1:
0.20
0.18
0.16
log [(V-V0)/(V-Vt)]
0.14
0.12
0.10
0.08
0.06
0.04
Slope=0.00495
-2
-1
k = 1.14 x 10 min
0.02
0.00
0
5
10
15
20
25
30
35
40
45
t / min
Prepared by Dr. S. Saha, Chemistry, BHU
4
Student 3:
vinf
V0sep
28
V0inst
v
16
16.2
t
vinf-vt
16.5
17.5
18.3
19.8
20.2
5
10
20
30
40
vinf-vos
11.5
10.5
9.7
8.2
7.8
vinf-v0ins log-sep
log-ins
12
11.8 0.018483 0.011184
0.057992 0.050693
0.09241 0.08511
0.165367 0.158068
0.187087 0.179787
0.20
0.20
V0 from reaction mixture
0.18
V0 separately determined
0.18
0.16
0.16
0.14
log [(V-V0)/(V-V0)]
log [(V-V0)/(V-V0)]
0.14
0.12
0.10
0.08
Slope=0.00464
-2
-1
k= 1.08x10 min
0.06
0.04
0.02
0.12
0.10
0.08
Slope=0.00495
-2
-1
k= 1.14x10 min
0.06
0.04
0.02
0.00
0
5
10
15
20
25
30
35
t/min
Prepared by Dr. S. Saha, Chemistry, BHU
40
45
0.00
0
5
10
15
20
25
30
t/min
5
35
40
45